Question

If f(x) and g(x) are differentiable functions in [0,1] such that f(0)=2=g(1), g(0)=0, f(1)=6 then there exists c, $0<c<1$ such that f '(c) =

• A. g '(c)
• B. -g '(c)
• C. 2g '(c)
• D. 3g '(c)

Answer 1

Answer: (C)

2g '(c)

We have,

$f\left(0\right)=2,$

$g\left(0\right)=0$

$f\left(1\right)=6$

$g\left(1\right)=2$

$f^{\prime }\left(c\right)=?$

We know that,

By mean value theorem

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$=\frac{6-2}{1}$

$f^{\prime }\left(c\right)=4$

Now,

$g^{\prime }\left(c\right)=\frac{g\left(1\right)-g\left(0\right)}{1-0}$

$=\frac{2-0}{1}$

$g^{\prime }\left(c\right)=2$

Hence, $f^{\prime }\left(c\right)=2.g^{\prime }\left(c\right)$

This is the answer.